I have been thinking about the Goldbach Conjecture, (G.C.), off and on for the last few years and decided to explore using an algebraic method of proof. This has the benefits, as you will see, that very little needs to be known about the distribution of the prime numbers in order to prove the conjecture is true. I will put all lemmas, corollaries, and theorems in bold for easy reading.

First, it is important to state that the Goldbach Conjecture says that every even number greater than 2 can be written as the sum of two prime numbers. The goal is to determine algebraically what conditions are needed for the Goldbach Conjecture to fail and show those conditions cannot be met. In order to proceed it is important to note that throughout this proof the variable, $a$, will always be assumed to be a natural number greater than 3 and non-prime. Also, the set of natural numbers, $mathbb{N}$, will include zero. $mathbb{P}$ will represent the set of prime numbers. With this in place everything is needed to begin.

If the Goldbach Conjecture were false for some non-prime $a in mathbb{N}$ with $a > 3$, then the closure property of the integers ensures for any prime $p_i < a$ there exists some $q_i in mathbb{N}$ where

$$2a = q_i + p_i tag{Equation 1}$$

with no $q_i$ being prime. With $a < forall q_i < 2a$, there must exist $alpha_1, alpha_2, dots, alpha_{pi(a)} in mathbb{N}$ allowing for

$$prod_{i = 1}^{pi(a)}q_i = prod_{i = 1}^{pi(a)}p_i^{alpha_i}. tag{Equation 2}$$ Conversely, if for each prime $p_i < a$ there is some $q_i in mathbb{N}$ where $2a = q_i + p_i$ and there exists $alpha_1, alpha_2, dots, alpha_{pi(a)} in mathbb{N}$ allowing for $prod_{i = 1}^{pi(a)}q_i = prod_{i = 1}^{pi(a)}p_i^{alpha_i}$, then no $q_i$ can be prime. This would provide a counter-example to the Goldbach Conjecture.

Therefore, following lemma is true.

$$textbf{The G.C. is false only if there exists some $a in mathbb{N}$ where $a > 3$ and Equations 1, 2 are satisfied.} tag{Lemma 1}$$

The next step is to try and find solutions to these sets of equations. Notice from Equation 1 that it follows for any primes $p_i, p_j < a$ there must exist non-prime $q_i, q_j$ where

$$q_j + p_j = 2a = q_i + p_i. tag{Equation 3}$$

If some $p_i$ existed where $p_i | q_i$ and $p_i | q_j$ in Equation 3, then $p_i | p_j$. Since both $p_i, p_j in mathbb{P}$, it is impossible for $p_i | p_j$ when $j neq i$. Therefore, an important corollary can be produced

$$textbf{if any $p_i | q_i$, then $p_i nmid q_j$ when $j neq i.$} tag{Corollary 1}$$

The crucial final step is to show that it is impossible for any $p_i|q_j$ when $j neq i.$ This will be done by asking about the truth value of the inverse of Corollary 1. This is given by the statement if $p_i nmid q_i$, then $p_i|q_j$ for some $jneq i$.

To begin, notice that under Equation 2, if any $p_i | q_j$ for some $j neq i$, then $p_i|p_i^{alpha_i}$. This allows for another corollary given below.

$$textbf{If any $p_i|q_j$ for some $j neq i$, then $alpha_i > 0.$} tag{Corollay 2}$$

It will now be shown that the inverse of Corollary 1 is never true when its antecedent is true. To begin, take the conjunction of the inverse of Corollary 1 and Corollary 2 below.

$$text{(if $p_i nmid q_i$, then $p_i|q_j$ for some $jneq i$)$; land ;$ (if any $p_i|q_j$ for some $j neq i$, then $alpha_i > 0)$}$$

The conjunction above allows for the following conclusion via transitivity.

$$text{if $p_inmid q_i$, then $alpha_i > 0.$}$$

The law of contraposition states that the truth value of a conditional shares the same truth value of its contrapositive, therefore, it follows below.

$$text{(if $p_inmid q_i$, then $alpha_i > 0.)$ $; equiv ;$ (if $alpha_i = 0$, then $p_i|q_i$}.)$$

However, it is impossible for $alpha_i = 0$ and $p_i|q_i$ since this would require the two equal products in Equation 2 to have differing prime factors. This leads to a contradiction, as it violates the Fundamental Theorem of Arithmetic. This means that if $p_inmid q_i$, then $alpha_i > 0$ is false whenever $p_inmid q_i$ which demonstrates the inverse of Corollary 1 must also be false whenever $p_i nmid q_i$. Since either $p_i$ is a divisor of $q_i$ or not, along with the truth of Corollary 1, there exists no scenario where $p_i | q_j$ when $j neq i.$

Since it is impossible for any $p_i | q_j$ when $j neq i$ solutions to Lemma 1 must be of the form $q_i = p_i^{alpha_i}$. A substitution into Equation 1 shows

$$text{$2a = p_i^{alpha_i} + p_i$ for all prime $p_i < a$}tag{Equation 4}$$

forcing all $p_i > a$ to be divisors of $2a$. Bertrand’s Postulate ensures $2a < a#$ for $a > 4$, showing solutions cannot exist when $a > 3$. Under Lemma 1 the Goldbach Conjecture is true.

A similar procedure can be used to show that every even number is the difference of two prime numbers. Also, using the weak version of the Goldbach Conjecture along with everything proven thus far, it is possible to prove the Polignac Conjecture. I do not wish to go over all of those steps here, so a link to my paper is given by https://osf.io/terdc/download. I have not been able to put this on the ARXIV since I have no university affiliation, and am honestly worrying about someone stealing the idea. At this point I have no other way to get this out to see if it has any merit and I have been sitting on it for years now hoping there would be another way. The journals I have submitted to have rejected it without any cause about 20 minutes after submitting it. If there is a mistake then I will accept that, but I cannot seem to find any. Thank you for any time that can be given. My email is in the paper.